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Evaluate $\frac{x^2 + x - 2}{2x^2 + x -3}$ when x = -1...

Evaluate $\frac{x^2 + x - 2}{2x^2 + x -3}$ when x = -1

A.

-2

B.

-1

C.

$-\frac{1}{2}$

D.

1

View answer & explanation

Correct answer is D

$\frac{x^2 + x - 2}{2x^2 + x - 3}$

= $\frac{x^2 + 2x - x - 2}{2x^2 + 3x - 2x - 3}$

= $\frac{x(x + 2) - 1(x + 2)}{x(2x + 3) - 1(2x + 3)}$

= $\frac{(x - 1)(x + 2)}{(x - 1)(2x + 3)}$

= $\frac{x + 2}{2x + 3}$

At x = -1,

= $\frac{-1 + 2}{2(-1) + 3}$

= $\frac{1}{1}$

= 1

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